How do you integrate int 1/(xsqrtx)dx?

Feb 16, 2017

The answer is $= - \frac{2}{\sqrt{x}} + C$

Explanation:

We use

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

$\sqrt{x} = {x}^{\frac{1}{2}}$

Therefore,

$\int \frac{\mathrm{dx}}{x \sqrt{x}}$

$= \int \frac{\mathrm{dx}}{{x}^{\frac{3}{2}}}$

$= \int {x}^{- \frac{3}{2}} \mathrm{dx}$

$= {x}^{- \frac{3}{2} + 1} / \left(- \frac{3}{2} + 1\right) + C$

$= - 2 {x}^{- \frac{1}{2}} + C$

$= - \frac{2}{\sqrt{x}} + C$

Feb 16, 2017

$- \frac{2}{\sqrt{x}} + C$

Explanation:

Use power rule to integrate as shown below: